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Why FVM for high-Re flows?

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Old   May 14, 1999, 05:49
Default Re: What is this flow that you mentioned?
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I fail to see how you can attach the meaning you seem to have done to those two sentences. To repeat myself, at the wall the fluid is not moving (stationary) and is therefore not performing work against the wall shear stress. However, in the boundary layer the fluid is moving and is performing work against (some) stress components.

I would like to bring this discussion to a close if we may.
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Old   May 14, 1999, 07:55
Default I feel a bit confused
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To say truth, a little more than a bit ....

I try to clarify myself thinking at the real world. For instance I could think at a wind tunnel (with a closed loop of flow, but I think that this can work with an open loop too, with a few arrangements) with a model in it. The model is fixed to the heart (as like as the tunnel) and, if we suppose the heart will not move for the action of the tunnel, it can't perform any work.

On the other side, the fluid will be activated by some blades, which are tranferring some energy to the fluid. So blades perform some work on the fluid (otherwise, why we need power to make them rotating?). The motor which activates blades applies generalized forces to the shaft and to the structure of the tunnel so that it is equilibrated.

Now I guess we can say that the structure of the tunnel will not apply any work to the heart, which will not move because of the action of the tunnel. So we can think at the tunnel itself as a good fixed reference frame for the study. In this reference, the only thing which performs any work are the blades which activate the fluid. This energy is transferred to the fluid, which, in turn, has to dissipate it anyhow, as there isn't any other moving surface to work on. This energy must be dissipateed because when the fluid comes back to the blades it has lost the Delta(E) received from them (Stationary condition).

Now let's go and simulate a piece (only the model >>or<< only the blades) of this tunnel. I stand in front of two different situations from the energetical point of view:

1) I have some non working surfaces (e.g. the model)

2) I have some working surfaces (blades)

Is there any difference in simulating these two cases?

If no, I can't understand the debate between Andy and Patrick Godon.

If yes, please tell me which ones, if you can.

Thanks, Marco
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Old   May 14, 1999, 09:32
Default Re: I feel a bit confused
John C. Chien
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(1). When you apply a force ( F ) to an object, you also receive a reaction force ( -F ) from the object. That is Newton's law of action and reaction. ( my understanding of the law ) (2). The work done (per uint time) can be defined as ( F x V ) where V is the speed of the object. (3). So, if you are applying force on a chair ( sitting on a chair ) and the chair is not moving ( V=0 ), then ( F x V )=0. In other word, when you are sitting on a chair for the whole day and the chair is not moving, you are not producing any work, that is , you are relaxing! You are not working. (4). If you decide to push a building from sunrise to sundown, then you will be exhausted at the end of the day.(energy dissipated) The work done (in one day)on the building is ( F x V ), which will be zero. And the whole day's work will be nothing. (5). In the above definition, I have used V (speed) instead of S (the distance moved ), because it is easier to understand. (when something moves, it must have a speed associated with it.) (6). I have discovered recently that the CFD code I am using has at least one big error in it, because two persons modified the code at two different occasions had two different definitions of a parameter, which are not consistent with the original definition. ( unless a code is documented line-by-line, definition-by-definition, subroutine-by-subroutine, it is not safe to touch the code. )
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Old   May 14, 1999, 13:22
Default Re: I feel a bit confused
John C. Chien
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(1). I am going to briefly describe the difference between two cases of simulation, one for the stationary model, and one for the moving blade. (2). For the stationary model, we solve governing equations written in stationary frame of reference. There will be a total pressure loss between the fluid entering and leaving the fixed model, that is ahead and behind the model because of the viscous effect between the fluid and the fixed model. The total pressure of the fluid leaving the model is lower....(3). For the moving blade, normally we solve governing equations in the moving frame of reference, in this case, moving with the blade (that is fixed on the blade). Mainly this is because it is much harder to solve the governing equations in fixed frame of reference with moving blade wall boundary.( the flow will be transient and you have to keep track of the blade wall location at any time step.) In the moving frame of reference, the velocity computed will be the relative velocity. And the relative velocity between the fluid and the blade at the blade wall will be zero,and both the fluid and the blade are moving in the stationary frame of reference. (4).When you compute the absolute total pressure (in fixed frame of reference) for the fluid leaving the moving blade, it will have contributions from two sources. One is related to the total pressure loss between the blade and the fluid ( viscous loss), and the other is the total pressure gain because of the work-done by the blade on the fluid. The force in this case is the integrated surface pressure force on both side of the blade. In the stationary frame of reference, the contact surface is moving (rotating) and there is a force acting at the contact surface ( the blade surface). So, there is work-done on the fluid by the moving blade. The net gain of the total pressure of fluid leaving the blade is positive, because the viscous loss is relatively small ( say, a few percent ). (5). For the closed-loop wind tunnel, in addition to the viscous loss between the fluid and the wall, heating and cooling are also required to maintain the test-section condition constant. Without cooling, the fluid temperature will increase continuously.
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